huda wrote:
In a room filled with 7 people, 4 people have exactly 1 sibling in the room and 3 people have exactly 2 siblings in the room. If two individuals are selected from the room at random, what is the probability that those two individuals are NOT siblings?
A. \(5/21\)
B. \(3/7\)
C. \(4/7\)
D. \(5/7\)
E. \(16/21\)
First we need to recognize that the given information tells us that the 7 people consist of:
- a sibling trio
- a sibling pair
- and another sibling pair
Let's use
counting techniques to answer this question
For this question, it's easier to find the complement.
So P(not siblings) = 1 -
P(they are siblings)P(they are siblings) = [# of ways to select 2 siblings] / [total # of ways to select 2 people]
# of ways to select 2 siblingsCase a) 2 siblings from the sibling trio: from these 3 siblings, we can select 2 siblings in 3C2 ways (3 ways)
Case b) 2 siblings from first sibling pair: we can select 2 siblings in 2C2 ways (1 way)
Case c) 2 siblings from second sibling pair: we can select 2 siblings in 2C2 ways (1 way)
So, total number of ways to select 2 siblings = 3+1+1 = 5
total # of ways to select 2 peopleWe have 7 people and we want to select 2 of them
We can accomplish this in 7C2 ways (21 ways)
So,
P(they are siblings) =
5/21This means P(
not siblings) = 1 -
5/21= 16/21
Answer: E
Cheers,
Brent